Y - 6 = 10(x - 2) hope this helps :)
Answer:
thanks
Step-by-step explanation:
Here are the steps required for Simplifying Radicals:
Step 1: Find the prime factorization of the number inside the radical. Start by dividing the number by the first prime number 2 and continue dividing by 2 until you get a decimal or remainder. Then divide by 3, 5, 7, etc. until the only numbers left are prime numbers. Also factor any variables inside the radical.
Step 2: Determine the index of the radical. The index tells you how many of a kind you need to put together to be able to move that number or variable from inside the radical to outside the radical. For example, if the index is 2 (a square root), then you need two of a kind to move from inside the radical to outside the radical. If the index is 3 (a cube root), then you need three of a kind to move from inside the radical to outside the radical.
Step 3: Move each group of numbers or variables from inside the radical to outside the radical. If there are nor enough numbers or variables to make a group of two, three, or whatever is needed, then leave those numbers or variables inside the radical. Notice that each group of numbers or variables gets written once when they move outside the radical because they are now one group.
Step 4: Simplify the expressions both inside and outside the radical by multiplying. Multiply all numbers and variables inside the radical together. Multiply all numbers and variables outside the radical together.
Shorter version:
Step 1: Find the prime factorization of the number inside the radical.
Step 2: Determine the index of the radical. In this case, the index is two because it is a square root, which means we need two of a kind.
Step 3: Move each group of numbers or variables from inside the radical to outside the radical. In this case, the pair of 2’s and 3’s moved outside the radical.
Step 4: Simplify the expressions both inside and outside the radical by multiplying.
since by the provided tickmarks WE ≅ ER, and QE ≅ ET, then both triangles are congruent by LL theorem, Leg Leg.
anyhow, since those legs are congruent, their hypotenuses must also be congruents, and we could use the HL theorem as well.
